## anonymous 5 years ago Find all critical points of f(z)=(3z^2-6)^2(1-5z)^4

1. anonymous

z = 2 and z = 1/5

2. anonymous

Take the derivative of f(z) with respect to z and set it equal to zero. The values of z that result will be critical points; don't know if you want to distinguish between different types though.

3. anonymous

I understand how to find them - I was having problems with setting the derivative to zero.

4. anonymous

Alright; give me a minute and I'll try it out. :)

5. anonymous

I've got f'(z)=6(z^2-2)(6z)(1-5z)^4-15(z^2-2)4(1-5z)^3 for the derivative. Look like what you got?

6. anonymous

When you differentiate, it should come out to this:$f'(z) = 6*(1-5z)^4*(3z^2-6)*2z-20*(1-5z)^3*(3x^2-6)^2$ Remember to not get sidetracked by careless errors; they're very prevalent in this problem.

7. anonymous

Sorry, accidentally stuck an x into the last factor when it was meant to be a z :P

8. anonymous

The image cuts off at (3x^2 - 6)...is that the end? Just making sure

9. anonymous

I get a pretty simple quadratic equation in the end..yep.

10. anonymous

Hmm..How did you get to the quadratic equation? Sorry I don't understand

11. anonymous

I just moved the entire right term (starting with -20*(1-5z)^3...) to the left hand side of the equation, and started simplifying. Things begin cancelling immediately and it breaks down into a quadratic equation (or so I believe)

12. anonymous

You set them equal to each other and simplify? How can they simplify if they are being multiplied to each other - I though you can't do that.

13. anonymous

No...I took that derivative and set it equal to zero, then when I moved the term beginning with -20 to the opposite side (where the 0 would be) I began cancelling common factors.

14. anonymous

Did you get 0 = -60z^2 +12z - 20 at some point?

15. anonymous

Did you get 0 = -60z^2 +12z - 20 at some point?

16. anonymous

Hm...let me show you what I did: $0 = 6*(1-5z)^4*(3z^2-6)* (2z) - 20 * (1-5z)^3*(3z^2-6)^2$$20 * (1-5z)^3*(3z^2-6)^2= 6*(1-5z)^4*(3z^2-6)* (2z)$(cancel like terms and move coefficients)$\frac{5}{3}*(3z^2-6) = (1-5z)*z$$5z^2-10=z-5z^2$$10z^2-z-10 = 0$ and you can solve after that.

17. anonymous

Ohhhhh...Okay. Thank you very much! I appreciate your help.

18. anonymous

Glad to help. :)

19. anonymous

The first person to answer this was incorrect, right? You should get some weird decimal numbers after using quadratic formula?